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interference fringes are observed by a byprism of refracting angle 1degree and refractive index 1.5 on a screen placed 100cm away . the wavelength of light used is 5890 angstrom . distance of source from prism is 20 cm . what is the fringe width

interference fringes are observed by a byprism of refracting angle 1degree and refractive index 1.5 on a screen placed 100cm away . the wavelength of light used is 5890 angstrom . distance of source from prism is 20 cm . what is the fringe width

Grade:12th pass

1 Answers

Deep Jyoti Rabha
15 Points
2 years ago
given, n=1.5     r=20cm   D=100+20    A=10
                                          =120cm        =0.0174 radian
 
\delta =(n-1)A
    = (1.5-1)*0.0174
    =0.0087266
d=2r\delta
    = 2*20*0.0087266
    = 0.349 cm
B=D\lambda /d
      = (120*5890*10-8)/0.349
      = 0.02024 cm
      = 0.202 mm

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